Exercise 41 Page 342

We are given that $\overline{AD}$ is perpendicular to $\overline{CE}.$ Also, we know that $\overline{CE}$ and $\overline{AD}$ are the medians of $△ABC.$ This means that $E$ and $D$ are the midpoints of $\overline{AB}$ and $\overline{CB}$ respectively. Let's mark these pieces of information on the given diagram.

As we can see, $△AOC$ is a right triangle. Thus, to find $AC$ we can use the Pythagorean Theorem. Although, first we need to find $CO$ and $AO.$ Let's do this!

Finding $CO$

We are going to use the fact that $\overline{CE}$ and $\overline{AD}$ are the medians of $△ABC.$ A point of intersection of triangle medians, which in our case is $O,$ is called a centroid. Let's recall what the Centroid Theorem states. According to this theorem, $CO$ is two thirds of the distance from $C$ to the midpoint $E.$ It is given that segment $\overline{CE}$ measures $9.$ By substituting this value into the above equation we can calculate $CO.$

Finding $AO$

Let's use the diagram again.

We can see that segment $\overline{CE}$ consists of $\overline{CO}$ and $\overline{OE}.$ By the Segment Addition Postulate, its measure is the sum of measures of $\overline{CO}$ and $\overline{OE}.$ It is given that $\overline{CE}$ measures $9$ and we have found that $\overline{CO}$ measures $6.$ By substituting these values into this equation, we can find the measure of $\overline{OE}.$ We also know that $\overline{AB}$ measures $10.$ Because $\overline{CE}$ is a median of $\overline{AB},$ segments $\overline{AE}$ and $\overline{EB}$ are congruent and have the same measure. Dividing $10$ by $2,$ we get that each of them measures $5.$ Let's now consider the triangle $△AOE.$ It is a right triangle, as $\overline{CE}$ and $\overline{AD}$ are perpendicular and form a right angle $\angle AOE.$ Hence, we can apply to it the Pythagorean Theorem. We know the values of $OE$ and $AE,$ so we can substitute $OE$ with $3$ and $AE$ with $5.$ Let's solve this equation and find $AO.$

Finding $AC$

Now that we know the measures of $\overline{CO}$ and $\overline{AO},$ we can find $AC.$

Let's use the Pythagorean Theorem, which applied to $△AOC$ has the following form. If we substitute $AO$ with $4$ and $CO$ with $6,$ we will get the equation where the only unknown is $AC.$ Let's solve it! Therefore, segment $\overline{AC}$ measures $2\sqrt{13}.$